Answer the following. (i) An object of 5 cm height is placed at a distance of 20 cm from the optical centre of a concave lens of focal length 18 cm. Calculate (1) image distance and (2) the magnification in this case. (ii) Compare the values of magnification obtained by a concave lens and a convex lens when both the lenses form virtual images.

ScienceCBSE Class 10 · Board exam 2023, set 31/5/1 · 3 marks · Short answer

CBSE Class 10 Science 2023: question 30(a)

Set in the CBSE Class 10 board exam of 2023, paper 31/5/1, section C, question 30(a), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q30(a) · 3 marks · Short answerLight – Reflection and Refraction

Answer the following.

(i)
An object of 5 cm height is placed at a distance of 20 cm from the optical centre of a concave lens of focal length 18 cm. Calculate (1) image distance and (2) the magnification in this case.
0 marks
(ii)
Compare the values of magnification obtained by a concave lens and a convex lens when both the lenses form virtual images.
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 31/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

By the sign convention, what is the sign of the focal length of a concave lens?

Hint 2

Use 1/v − 1/u = 1/f and rearrange for 1/v.

Hint 3

A virtual image from a convex lens is enlarged, and from a concave lens it is reduced. How does that show up in m?

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Given: $h = 5$ cm, $u = - 20$ cm, $f = - 18$ cm (concave lens). Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
½ mark
(1) $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 18} + \frac{1}{- 20} = - \frac{10 + 9}{180} = - \frac{19}{180}$ , so $v = - \frac{180}{19} \approx - 9.47$ cm. The image is virtual and on the same side as the object.
½ mark
(2) Magnification: $m = \frac{v}{u}$
½ mark
$m = \frac{- 9.47}{- 20} \approx 0.47$ . This is positive and less than 1, so the image is erect and diminished (image height ≈ 2.37 cm).
½ mark
(ii) Convex lens forming a virtual image: the image is erect and magnified, so $m$ is positive and ** $m > 1$ **.
½ mark
Concave lens: the virtual image is always erect and diminished, so $m$ is positive and ** $m < 1$ **.
½ mark

Answer: (i) v ≈ −9.47 cm, m ≈ +0.47; (ii) virtual image: convex lens m > 1, concave lens m < 1 (both positive)

Where the marks are

Uses the lens formula with correct signs (½ mark)
Obtains v = −9.47 cm (½ mark)
Uses m = v/u (½ mark)
Obtains m = 0.47 (½ mark)
Convex lens virtual image: m > 1 (½ mark)
Concave lens: m < 1 (½ mark)

Revise the chapter

Chapter: Light – Reflection and Refraction (notes, formulas, where marks are lost and every past question)
Topic: Lens Formula, Magnification and Power of a Lens

Practise this chapterBoard-style questions from Light – Reflection and Refraction, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 31/5/1: every question, by section.

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CBSE Class 10 Science 2023: question 30(a)

Hints

Step-by-step solution

Revise the chapter

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